Physics High Jump Question

1. The problem statement, all variables and given/known data
If a high jumper can clear a 2m bar on Earth, how high is the bar he can clear on the moon assuming it is done under the same conditions as on the Earth. Acceleration due to gravity on the moon is 1.7 m/s^2.

2. Relevant equations
GPE = mgh

3. The attempt at a solution
I set mgh on Earth equal to mgh on the moon as the athlete exerts the same starting energy transfer. I got a factor of 5.8 giving the bar on the moon at 11.6m. However, the mark scheme says the athlete lifts his COM 1m on Earth. I have no idea how you are suppose to know this? I get that the COM goes under the bar but what makes it half the height of the bar? This is a proper past exam question and what I have put in the question is all they give you! Please help. Thanks.

Staff: Mentor

However, the mark scheme says the athlete lifts his COM 1m on Earth. I have no idea how you are suppose to know this?

Good question! Perhaps you supposed to 'estimate' the height of the athlete's COM as being about 1 m off the ground when he leaves the ground. (Not cool for an exam question, unless it's free response and any reasonable estimate is OK.)

Good question! Perhaps you supposed to 'estimate' the height of the athlete's COM as being about 1 m off the ground when he leaves the ground. (Not cool for an exam question, unless it's free response and any reasonable estimate is OK.)

Phew! Glad I wasn't missing something. I have attached the paper and mark scheme - it's question 3b (ii). It doesn't even say the 1m COM is an estimate - it states it as though it is a fact! Thanks

This is a common blunder.
A high jumper does not arrive at the bar in a vertical position then jump. Would not go very high that way.
The jump starts with a run up. Approaching the bar, the jumper leans back and bends the legs. A combination of the forward momentum and leg extension from that low position produces the upward velocity needed. So the mass centre has to lifted from rather lower than 1m.